15. Derivatives of Trigonometric Functions In this - WebAssignкод для вставки
15. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 47 15. Derivatives of Trigonometric Functions In this section, we show how to compute the derivatives of trigonometric functions. First, we compute (sin x) . This will be a somewhat lengthy procedure, due to the fact that this is the п¬Ѓrst trigonometric function we will diп¬Ђerentiate and we will have to apply new methods. However, once we know the derivatives of sin x and cos x, it will be much simpler to deduce the derivatives of other trigonometric functions, since those functions can be obtained from sin and cos, and then the various diп¬Ђerentiation rules can be used. Theorem 16. We have (sin x) = cos x. Proof. Recall the identity sin(a + b) = sin a cos b + sin b cos a. We have sin(x + h) в€’ sin x (sin x) = lim hв†’0 h sin x cos h + sin h cos x в€’ sin x = lim hв†’0 h (cos h) в€’ 1 sin h cos x + = lim sin x hв†’0 h h (cos h) в€’ 1 sin h = sin x lim + cos x lim . hв†’0 hв†’0 h h Note that as, h approaches 0, we certainly have limhв†’0 sin x = sin x and limhв†’0 cos x = cos x, since these functions do not even depend on h. There remains the task of computing the two nontrivial limits (cos h) в€’ 1 sin h and lim . hв†’0 h h We will carry out this task in two lemmas. lim hв†’0 Lemma 1. We have sin h = 1. hв†’0 h Proof. Let us consider a circle with unit radius and a regular ngon whose center is at the center O of the circle and whose n vertices are all on the unit circle. Then the area of the circle is ПЂ, and the area of the n-gon is n В· 12 В· sin О±, where О± = 2ПЂ/n is the angle AOB, with A and B being adjacent vertices of our n-gon. Considering just 1/n of both the circle and the n-gon, we see that the area of the triangle AOB is (sin О±)/2, and the area of 1/n of the lim 48 3. RULES OF DIFFERENTIATION circle bordered by the lines AO, BO, and the arc AB is ПЂ В· О±/(2ПЂ) = О±/2. So the ratio of the two areas is (sin О±)/2 sin О± . = О±/2 О± On the other hand, as n gets larger and larger, О± gets smaller and smaller, while the area of the n-gon gets closer and closer to the area of the circle. Hence, their ratio, sin О±/О±, will get arbitrarily close to 1 and stay arbitrarily close to 1. вњ· Lemma 2. The equality (3.4) lim hв†’0 (cos h) в€’ 1 =0 h holds. Proof. We will manipulate the expression ((cos h) в€’ 1)/h so that we can use the result of Lemma 1. First, we multiply both the numerator and the denominator by cos h + 1 to get (cos2 h) в€’ 1 в€’ sin2 h (cos h) в€’ 1 = = . h h(1 + cos h) h(1 + cos h) Therefore, we have (cos h) в€’ 1 sin2 h = в€’ lim hв†’0 hв†’0 h(1 + cos h) h sin h sin h В· = в€’ lim hв†’0 h 1 + cos h sin h sin h = в€’ lim В· lim hв†’0 h hв†’0 1 + cos h = (в€’1) В· 0 = 0. lim вњ· We can now п¬Ѓnish the proof of Theorem 16. At the end of the п¬Ѓrst displayed chain of equations in that proof, we saw that (cos h) в€’ 1 sin h + cos x lim . hв†’0 hв†’0 h h The previous two lemmas showed that, on the right-hand side, the п¬Ѓrst limit is 0 and the second limit is 1, so (sin x) = cos x as claimed. вњ· (sin x) = sin x lim The following theorem can be proved by very similar methods. Theorem 17. The equality (cos x) = в€’ sin x holds. 16. THE CHAIN RULE 49 You are asked to prove this theorem in Exercise 15.1(1). Now that we have the derivatives of sin and cos, the derivatives of other trigonometric functions can be obtained by simply using the quotient rule. The next theorem shows an example of this. Theorem 18. We have (tan x) = sec2 x. Proof. Note that tan x = sin x/cos x, so we can apply the quotient rule. This leads to sin x (tan x) = cos x cos x В· (sin x) в€’ sin x(cos x) = cos2 x cosx + sin2 x = cos2 x 1 = cos2 x вњ· = sec2 x. The derivatives of the other three trigonometric functions are given in the exercises. 15.1. Exercises. (1) (2) (3) (4) (5) (6) Prove that (cos x) = в€’ sin x. Prove that (cot x) = в€’ csc2 x. Prove that (csc x) = в€’ csc x В· cot x. Prove that (sec x) = sec x tan x. Let h(x) = ex cos x. Find h (x). Let h(x) = ex / sin x. Find h (x). 16. The Chain Rule 16.1. The Derivative of the Composition of Two Functions. In previous sections, we learned how to compute the derivative of the sum, diп¬Ђerence, product, and quotient of two functions. We still do not know how to compute the derivative of the composition of functions, such as в€љ h(x) = sin(3x), t(x) = x2 + 1, or r(x) = e4x . In this section, we will learn a rule, called the chain rule, that applies in these situations. Theorem 19 (Chain Rule). Let h(x) = f (g(x)), where g is diп¬Ђerentiable at x and f is diп¬Ђerentiable at g(x). Then h is diп¬Ђerentiable at x, and we have h (x) = f (g(x))g (x).